# Inequality of real numbers with exponent

For $$a,b>0$$ are two real numbers and $$p\geq 1$$. Is the following inequality true $$|a^p-b^p|\leq|a-b|^p\;\;?$$

• 5=|3^2-2^2| > |3-2|^2=1 – ALG Oct 26 '18 at 21:10

No, let $$a=10,b=20,p=2$$.
Then $$|a^p-b^p|=|100-400|=300>100=|10-20|^2=|a-b|^p$$.
Let $$b\geq a$$ then $$\>b=a+\delta\>$$ and $$\>\delta>0$$ $$\\|a^p-b^p|\leq|a-b|^p \\(a+\delta)^p-a^p\leq\delta^p \\(a+\delta)^p\leq a^p+\delta^p$$ but $$a>0$$ and $$\delta\geq0 => (a+\delta)^p=a^p+\delta^p+\delta*x\>(x\geq0)($$ Binomial theorem$$)=>$$ $$\\a^p+\delta^p\geq(a+\delta)^p=a^p+\delta^p+\delta*x\geq a^p+\delta^p$$ $$=>\delta*x=0=>\delta=0\>$$ or $$\>x=0=>$$ $$\\|a^p-b^p|\leq|a-b|^p$$ if and only if $$a=b$$ or $$p=1$$
$$\left|1^2-\frac1{2^2}\right|\le\left|1-\frac12\right|^2\;?$$